Question: Determine how many solutions exist for the system of equations. ${4x-y = 10}$ ${6x+y = 1}$
Answer: Convert both equations to slope-intercept form: ${4x-y = 10}$ $4x{-4x} - y = 10{-4x}$ $-y = 10-4x$ $y = -10+4x$ ${y = 4x-10}$ ${6x+y = 1}$ $6x{-6x} + y = 1{-6x}$ $y = 1-6x$ ${y = -6x+1}$ Just by looking at both equations in slope-intercept form, what can you determine? ${y = 4x-10}$ ${y = -6x+1}$ The linear equations have different slopes. ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$ ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$ When two equations have different slopes, the lines will intersect once with one solution.